已知函数满足微分方程xy'=yln(yIx),且x=1时,Y=e^2是当x=-1时,Y=

Question

已知函数满足微分方程xy'=yln(yIx),且x=1时,Y=e^2是当x=-1时,Y=

Answer 1

xy' = yln(y/x)
令y = xv,y' = v + x · dv/dx = v + x · v'
v + x · v' = v · ln(v)
v' = (vln(v) - v)/x
∫ dv/[v(ln(v) - 1)] = ∫ 1/x dx
∫ d(ln(v) - 1)/[ln(v) - 1] = ∫ 1/x dx
ln[ln(v) - 1] = ln(x) + lnC = ln(Cx)
ln(v) = Cx + 1
v = e^(Cx + 1)
y/x = e^(Cx + 1)
y = xe^(Cx + 1)
当x = 1,y = e²
e² = e^(C + 1) => 2 = C + 1 => C = 1
∴通解y = xe^(x + 1)
当x = - 1,y = - 1 · e^(- 1 + 1)
= - 1